An electron is moving horizontally enters a region of constant electric field fi

Question

An electron is moving horizontally enters a region of constant electric field field produced by a parallel plate capacitor with a positive plate on the bottom and the negative plate on the top. The electric field is directed perfectly vertically. The electron enters the field with a speed of 4.55 x 10^6 m/s and exits the field 0.618 cm lower than it entered.

a) What is the magnitude of the electric field produced by the capacitor?

b) What is the velocity of the electron when it exits the capacitor's field?

c) Did you include the force of gravity in this problem? Why doesn't it matter?

Explanation / Answer

Mass of Electron = 9.11 * 10^-31 Kg

The time the electron is between the plates = Distance between the plates / Speed of Electron
The time the electron is between the plates = 0.025 m / 4.55 * 10^6 m/s = 5.5 * 10^-9 s
We know , F = q * E

Now, the acceleration of the electron (a) = F/m = E*q/m
y = 1/2*a*t^2 = 1/2* (E*q/m )* t^2

Thereforw E = 2*y*m/(q*t^2)
E = (2 * 0.618x10^-2 * 9.11x10^-31 ) / (1.60x10^-19* (5.5 * 10^-9 ) ^2 ) = 2326.43 N/C

Magnitude of the electric field produced E = 2326.43 N/C

B)
   vx doesn't change = 4.55 * 10^6 m/s
vy = a*t = (Eq/m) * t = (2326.43 *1.60x10^-19 /9.11x10^-31) * 5.5 * 10^-9 = 2.267 * 10^6m/s
Speed = sqrt(( 4.55 * 10^6 )^2 + (2.267 * 10^6)^2) = 5.08 * 10^6m/s

velocity of the electron when it exits the capacitor's field = 5.08 * 10^6m/s

C)
No , It doesn't matter because the mass of electron is Negligeble 9.11 * 10^-31 Kg , therefore force of Gravity is also very very small and can be neglected.

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